Should be 100
4C3 + 4C1*6C2 + 4C2*6C1 = 100

This is a combination problem. The best approach is to find the total possible groups (10C3 = 120), and subtract the number of groups with no senior partners ( consists of all junior partners so 6C3=20)

The answer is 120-20 = 100, B.

The longer approach would be to find out how many groups are possible with 1 senior partner, 2 senior partners, and 3 senior partners and add these.

1 senior partner: 4C1 * 6C2 = 4 * 15 = 60
2 senior partner: 4C2 * 6C1 = 6 * 6 = 36
3 senior partner: 4C3 * 6C0 = 4 * 1 = 4

Note that those sum to 100.

I would always advise to use the second approach for Permutation and Combination problems if one is not comfortable doing them. The first approach is neat and time saving. however they are for advanced users who can think one step ahead. Second approach is methodical and once you master the step by step approach the first approach logically follows.

Rahul

You are right. This is a combinations question. The groups are different based on the members in the group rather than how they are arranged. Let's consider Seniors as 'S' & Juniors as 'J'. The different groups are:

SJJ - 4c1*6c2 = 4*15 = 60
SSJ - 4c2*6c1 = 6*6 = 36
SSS - 4c3 = 4

So, the number of groups possible are = 60+36+4 = 100

If the groups also depend on the arrangement of the members, then the number of groups can be determined using both combinations & permutation.

SJJ - 4c1*6c2*3 = 4*15*3 = 180 (3 - number of ways the group can be arranged, i.e. SJJ, JSJ, JJS)
SSJ - 4c2*6c1*3 = 4*15*3 = 108 (3 - number of ways the group can be arranged, i.e. SSJ, SJS, JSS)
SSS = 4c3 = 4

So, the number of groups possible are = 180+108+4 = 292ways.

Hope this helps.